11.machines constrained flow shop scheduling processing time, setup time each associated with probabilities including job block criteria

Journal of Energy Technologies and Policy www.iiste.org
ISSN 2224-3232 (Paper) ISSN 2225-0573 (Online)
Vol.1, No.3, 2011

Solving n-Jobs, 3-Machines Constrained Flow Shop Scheduling
Processing Time, Setup Time Each Associated with
Probabilities Including Job-Block Criteria
Sameer Sharma1
Department of Mathematics, Maharishi Markandeshwar University, Mullana, Ambala, India
* E-mail of the corresponding author: samsharma31@yahoo.com

Abstract
This paper is pertain to heuristic technique for n-jobs, 3-machines flowshop scheduling problem in which
processing times and setup times are associated with their respective probabilities involving transportation
time, break down interval and job block criteria is taken in to account. Further jobs are attached with
weights to indicate their relative importance. The proposed method is very easy to understand and also
provide an important tool for decision makers. A numerical illustration followed by a computer programme
is also given to clarify the algorithm.
Keywords: Flow shop scheduling, Processing time, Setup time, Transportation time, Break down, Weights
of job, Job block.

1. Introduction
During the last 30 years, the flow shop sequencing problems has been the center of attention of many
researchers. Since Johnson had proposed optimal two and three stage production schedules, many heuristics
approaches have been suggested to solve the various problems. The flow shop scheduling problem is a
production scheduling problem in which each of the n jobs (tasks) must be processed in the same sequence
on each one of m machines (processors). The scheduling problem practically depends upon the important
factors namely, Job transportation which includes loading time, moving time and unloading time etc.,
Weightage of job which represents the relative importance of one job over another and Breakdown of
machine which is due to failure of electric current or due to non supply of raw material or any other
technical interruptions. The majority of scheduling research assumes setup which includes work to prepare
the machine as negligible or part of the processing time while this assumption adversely affects solution
quality for many applications which require explicit treatment of setup. Johnson (1954) proposed the well
known Johnson’s rule in the two stage flow shop scheduling problem. Yoshida & Hitomi (1979) further
considered the problem with setup times. The work was developed by Belman (1956), Maggu & Das
(1977), Miyazaki & Nishiyama (1980), Nawaz et al (1983), Singh (1985), Chandramouli (2005), Belwal
& Mittal (2008), Khodadadi (2008), Pandian & Rajendran (2010), Gupta & Sharma (2011) by considering
various parameters.
Gupta, Sharma & seema (2011) studied a n x 3 flowshop scheduling problem, processing time associated
with probabilities involving transportation time, breakdown interval, Weightage of jobs and job block
criteria. This paper is an attempt to extend the study made by Gupta & Sharma (2011) by introducing the
concept of independent setup time with their corresponding probabilities. We have obtained an algorithm
which minimize the total elapsed time whenever men weighted production flow time is taken into
consideration.

2. Practical Situation
Many applied and experimental situations exist in our day-to-day working in factories and industrial
production concerns etc. The practical situation may be taken in a paper mill, sugar factory and oil refinery
etc. where various qualities of paper, sugar and oil are produced with relative importance i.e. weight in jobs.
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In many manufacturing companies different jobs are processed on various machines. These jobs are
required to process in a machine shop A, B, C, ---- in a specified order. When the machines on which jobs
are to be processed are planted at different places, the transportation time (which includes loading time,
moving time and unloading time etc.) has a significant role in production concern. Setup includes work to
prepare the machine, process or bench for product parts or the cycle. This includes obtaining tools,
positioning work-in-process material, return tooling, cleaning up, setting the required jigs and fixtures,
adjusting tools and inspecting material and hence significant. The break down of the machines (due to delay
in material, changes in release and tails date, tool unavailability, failure of electric current, the shift pattern
of the facility, fluctuation in processing times, some technical interruption etc.) have significant role in the
production concern. The idea of job block has practical significance to create a balance between a cost of
providing priority in service to the customer and cost of giving service with non priority, .i.e. how much is
to be charged from the priority customer(s) as compared to non priority customer(s).

3. Notations
S : Sequence of jobs 1, 2, 3… n
Sk : Sequence obtained by applying Johnson’s procedure, k = 1, 2, 3, -------
Mj : Machine j, j= 1, 2, 3
M : Minimum makespan
aij : Processing time of ith job on machine Mj
pij : Probability associated to the processing time aij
sij : Set up time of ith job on machine Mj
qij : Probability associated to the set up time sij
Aij : Expected processing time of ith job on machine Mj
Sij : Expected set up time of ith job on machine Mj
β : Equivalent job for job – block
'
A : Expected processing time of ith job after break-down effect on jth machine
ij
Iij(Sk): Idle time of machine Mj for job i in the sequence Sk
Ti , j →k : Transportation time of ith job from jth machine to kth machine
wi : weight assigned to ith job
L : Length of break down interval.

4. Problem Formulation
Let some job i (i = 1,2,……..,n) is to be processed on three machines Mj ( j = 1,2,3). Let aij be the
processing time of ith job on jth machine with probabilities pij and sij be the setup time of ith job on jth
machine with probabilities qij. Let Ti,j→k be the transportation time of ith job from jth machine to kth machine.
Let wi be the weights assigned to the ith job. Our aim is to find a sequence {Sk } of the jobs which minimize
total elapsed time, and weighted mean-flow times whenever mean weighted production flow time is taken
into consideration.

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The mathematical model of the problem in matrix form can be stated as:

Job Machine A Ti ,1→2 Machine B Ti,2→3 Machine C Weight
s s of
jobs
i ai1 pi1 si1 qi1 ai 2 pi 2 si2 qi2 ai 3 pi 3 si3 qi3 wi

1 a11 p11 s1 q1 T1,1→2 a12 p12 s1 q1 T1,2→3 a13 p13 s1 q1 w1
2 a21 p21 1 1 T2,1→2 a22 2 2 T2,2→3 a23 p23 3 3 w2
3 s2 q2 a32 p22 s2 q2 s2 q2 w3
4 a31 p31 1 1 T3,1→2 a42
2 2 T3,2→3 a33 p33 3 3
w4
a41 s3 q3 p32 s3 q3 s3 q3
- - -
- p41 1 1 T4,1→2 2 2 T4,2→3 a43 p43 3 3
n s4 q4 an 2 p42 s4 q4 s4 q4 wn
an1 1 1 2 2 3 3
- - - - -
pn1 - - Tn,1→2 - - - Tn,2→3 an 3 pn3 - -
sn qn pn 2 sn qn sn qn
1 1 2 2 3 3

Table 1
5. Algorithm
Step 1: Calculate the expected processing times and expected set up times as follows
Aij = aij × pij and Sij = sij × qij ∀i, j =1,2,3
Step 2: Check the condition
Either Max {Ai1 + Ti,1→2 – Si2} ≥ Min{Ai2 + Ti,1→2 – Si1}
or Max{Ai3 + Ti,2→3 – Si2} ≥ Min{Ai2 + Ti,2→3 – Si3} or both for all i
If the conditions are satisfied then go to step 3, else the data is not in the standard form.
Step 3: Introduce the two fictitious machines G and H with processing times Gi and Hi as defined below:
Gi = Ai1 − Ai 2 − Ti ,1→ 2 − Ti ,2→3 − max( Si1 , Si 2 ) and H i = Ai3 − Ai 2 − Ti ,1→ 2 − Ti ,2→3 + Si3 .
Step 4: Compute Minimum ( Gi ,Hi)
If Min (Gi , Hi)=Gi then define Gi' =Gi + wi and H i' =Hi .
If Min (Gi , Hi)=Hi then define Gi' =Gi and H i' =Hi+ wi .
Step 5: Define a new reduced problem with Gi'' and H i'' where
Gi'' = Gi' wi , H i'' = H i' wi ∀i = 1, 2,3....., n
Step 6: Find the expected processing time of job block β = (k, m) on fictitious machines G & H using
equivalent job block criterion given by Maggu & Das (1977). Find G "β and H "β using
G "β = G "k + G "m − min(G "m , H "k ) and H "β = H "k + H "m − min(G "m , H "k )
Step7: Define new reduced problem with processing time Gi'' & H i'' as defined in step 5 and replace
job block β = ( k, m) by a single equivalent job β with processing times G "β & H "β as defined in step
6
Step 8: Using Johnson’s procedure, obtain all sequences Sk having minimum elapsed time. Let these be S1,
S2,....,Sr
Step 9: Prepare In-Out tables for the sequences S1 , S 2 ,.........., S r obtained in step 8. Let the mean flow

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time is minimum for the sequence Sk. Now, read the effect of break down interval (a, b) on different jobs
on the lines of Singh T.P. (1985) for the sequence Sk.
'
Step 10: Form a modified problem with processing time Aij ; i = 1,2,3,….,n; j= 1, 2,3
If the break down interval (a, b) has effect on job i then
Aij = Aij + L ; Where L = b – a, the length of break-down interval
'

If the break-down interval (a, b) has no effect on ith job then Aij = Aij .
'

Step 11: Repeat the procedure to get the optimal sequence for the modified scheduling problem using steps
3 to step 9. Determine the total elapsed time.
Step 12: Find the performance measure studied in weighted mean flow time defined by
n n
F = ∑ wi fi th
∑ wi , where fi is flow time of i job.
i =1 i =1

6. Programme
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<process.h>
#include<math.h>

int n,j;
float a1[16],b1[16],c1[16],a11[16],b11[16],c11[16],g[16],h[16],T12[16],T23[16],s11[16],s22[16],s33[16];
float macha[16],machb[16],machc[16],macha1[16],machb1[16],machc1[16],maxs[16];
int f=1;float minval,minv,maxv1[16],maxv2[16],minv1;float w[16];
int group[2];//variables to store two job blocks
int bd1,bd2;// Breakdown interval
float gbeta=0.0,hbeta=0.0;float gbeta1=0.0,hbeta1=0.0;
void main()
{
clrscr();
int a[16],b[16],c[16],j[16],s1[16],s2[16],s3[16];float p[16],q[16],r[16],x[16],t1[16],u[16];
cout<<"How many Jobs (<=15) : ";cin>>n;
if(n<1 || n>15)
{cout<<endl<<"Wrong input, No. of jobs should be less than 15..n Exitting";getch();exit(0);}
for(int i=1;i<=n;i++){j[i]=i;
cout<<"nEnter the processing time, set up time and the probabilities of "<<i<<" job for machine A and
Transportation time from Machine A to B : ";cin>>a[i]>>p[i]>>s1[i]>>x[i]>>T12[i];
cout<<"nEnter the processing time, setup time and the probabilities of "<<i<<" job for machine B and
Transportation time from Machine B to C : ";cin>>b[i]>>q[i]>>s2[i]>>t1[i]>>T23[i];
cout<<"nEnter the processing time and its probability of "<<i<<"job for machine C:
";cin>>c[i]>>r[i]>>s3[i]>>u[i];
cout<<"nEnter the weightage of "<<i<<"job:";cin>>w[i];
//Calculate the expected processing & setup times of the jobs for the machines:
a1[i] = a[i]*p[i];b1[i] = b[i]*q[i];c1[i] = c[i]*r[i];s11[i]=s1[i]*x[i]; s22[i]= s2[i]*t1[i]; s33[i]= s3[i]*u[i];}
cout<<endl<<"Expected processing time of machine A, B and C with weightage: n";
for(i=1;i<=n;i++)
{cout<<j[i]<<"t"<<a1[i]<<"t"<<s11[i]<<"t"<<T12[i]<<"t"<<b1[i]<<"t"<<s22[i]<<"t"<<T23[i]<<"t"<
<c1[i]<<"t"<<s33[i]<<"t"<<w[i];cout<<endl;}
cout<<"nEnter the two breakdown interval:";cin>>bd1>>bd2;
//Findinglargest in a1
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float maxa1;maxa1=a1[1]+T12[1]-s22[1];
for(i=2;i<n;i++)
{if((a1[i]+T12[i]-s22[i])>maxa1)maxa1=a1[i]+T12[i]-s22[i];}
//For finding smallest in b1
float minb1;minb1=b1[1]+T12[1]-s11[1];
for(i=2;i<n;i++)
{if(b1[i]+T12[i]-s11[i]<minb1)minb1=b1[i]+T12[i]-s11[i];}
float minb2;minb2=b1[1]+T23[1]-s33[i];
for(i=2;i<n;i++)
{if((b1[i]+T23[i]-s33[i])<minb2)minb2=b1[i]+T23[i]-s33[i];}
//Finding largest in c1
float maxc1;maxc1=c1[1]+T23[1]-s22[i];
for(i=2;i<n;i++)
{if((c1[i]+T23[i]-s22[i])>maxc1)maxc1=c1[i]+T23[i]-s22[i];}
for(i=1;i<=n;i++)
{if(s11[i]>s22[i]){maxs[i]=s11[i];}
else {maxs[i]=s22[i];}}
if(maxa1>=minb1||maxc1>=minb2)
{g[i]=fabs((a1[i]-T12[i]-b1[i]-T23[i]-maxs[i]));h[i]=fabs((c1[i]-T12[i]-b1[i]-T23[i]+s33[i]));}
else
{cout<<"n data is not in Standard Form...nExitting";getch();exit(0);}
for(i=1;i<=n;i++)
{g[i]=fabs(a1[i]-T12[i]-b1[i]-T23[i]-maxs[i]);h[i]=fabs(c1[i]-T12[i]-b1[i]-T23[i]+s33[i]);}
cout<<endl<<"Expected processing time for two fictious machines G and H: n";
for(i=1;i<=n;i++)
{cout<<endl;cout<<j[i]<<"t"<<g[i]<<"t"<<h[i]<<"t"<<w[i];cout<<endl;}
//To find minimum of G & H
float g1[16],h1[16];
for (i=1;i<=n;i++)if(g[i]<=h[i]){g1[i]=g[i]+w[i];h1[i]=h[i];}
else{g1[i]=g[i];h1[i]=h[i]+w[i];}
float g2[16],h2[16];
for(i=1;i<=n;i++)
{g2[i]=g1[i]/w[i];h2[i]=h1[i]/w[i];}
cout<<endl<<endl<<"displaying original scheduling table"<<endl;
for(i=1;i<=n;i++)
{cout<<j[i]<<"t"<<g2[i]<<"t"<<h2[i]<<endl;}
cout<<"nEnter the two job blocks(two numbers from 1 to "<<n<<"):"; cin>>group[0]>>group[1];
//calculate G_Beta and H_Beta
if(g2[group[1]]<h2[group[0]])
{minv=g2[group[1]];}
else{minv=h2[group[0]];}
gbeta=g2[group[0]]+g2[group[1]]-minv;hbeta=h2[group[0]]+h2[group[1]]-minv;
cout<<endl<<endl<<"G_Beta="<<gbeta;cout<<endl<<"H_Beta="<<hbeta;
int j1[16];float g13[16],h13[16];
for(i=1;i<=n;i++)
{if(j[i]==group[0]||j[i]==group[1]){f--;}
else{j1[f]=j[i];}f++;}
j1[n-1]=17;
for(i=1;i<=n-2;i++)
{g13[i]=g2[j1[i]];h13[i]=h2[j1[i]];}
g13[n-1]=gbeta;h13[n-1]=hbeta;
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for(i=1;i<=n-1;i++)
{cout<<j1[i]<<"t"<<g13[i]<<"t"<<h13[i]<<endl;}
float mingh[16];char ch[16];
for(i=1;i<=n-1;i++)
{if(g13[i]<h13[i])
{mingh[i]=g13[i];ch[i]='g';}
else {mingh[i]=h13[i];ch[i]='h'; }}
for(i=1;i<=n-1;i++)

for(int j=1;j<=n-1;j++)
{if(mingh[i]<mingh[j])
{float temp=mingh[i]; int temp1=j1[i]; char d=ch[i];mingh[i]=mingh[j]; j1[i]=j1[j]; ch[i]=ch[j];
mingh[j]=temp; j1[j]=temp1; ch[j]=d;}}
// calculate beta scheduling
float sbeta[16];int t=1,s=0;
for(i=1;i<=n-1;i++)
{if(ch[i]=='h'){ sbeta[(n-s-1)]=j1[i];s++;}
else if(ch[i]=='g'){sbeta[t]=j1[i];t++;}}
int arr1[16], m=1;cout<<endl<<endl<<"Job Scheduling:"<<"t";
for(i=1;i<=n-1;i++)
{if(sbeta[i]==17){arr1[m]=group[0];arr1[m+1]=group[1];
cout<<group[0]<<" "<<group[1]<<" ";m=m+2;continue;}
else {cout<<sbeta[i]<<" ";arr1[m]=sbeta[i];m++;}}
//calculating total computation sequence
float time=0.0,macha11[16];macha[1]=time+a1[arr1[1]];
for(i=2;i<=n;i++)
{macha11[i]=macha[i-1]+s11[arr1[i-1]];macha[i]=macha11[i]+a1[arr1[i]];}
machb[1]=macha[1]+b1[arr1[1]]+T12[arr1[1]];
for(i=2;i<=n;i++)
{if((machb[i-1]+s22[arr1[i-1]])>(macha[i]+T12[arr1[i]]))maxv1[i]=machb[i-1]+s22[arr1[i-1]];
else maxv1[i]=macha[i]+T12[arr1[i]];machb[i]=maxv1[i]+b1[arr1[i]];}
machc[1]=machb[1]+c1[arr1[1]]+T23[arr1[1]];
for(i=2;i<=n;i++)
{if((machc[i-1]+s33[arr1[i-1]])>(machb[i]+T23[arr1[i]]))
maxv2[i]=machc[i-1]+s33[arr1[i-1]];
else
maxv2[i]=machb[i]+T23[arr1[i]];machc[i]=maxv2[i]+c1[arr1[i]];}
cout<<endl<<endl<<"In-Out Table is:"<<endl<<endl;
cout<<"Jobs"<<"t"<<"Machine M1"<<"t"<<"t"<<"Machine M2" <<"t"<<"t"<<"Machine M3"<<endl;
cout<<arr1[1]<<"t"<<time<<"--"<<macha[1]<<"
t"<<"t"<<macha[1]+T12[arr1[1]]<<"--"<<machb[1]<<"
t"<<"t"<<machb[1]+T23[arr1[1]]<<"--"<<machc[1]<<endl;
if((time<=bd1 && macha[1]<=bd1)||(time>=bd2 && macha[1]>=bd2))
{a1[arr1[1]]=a1[arr1[1]];}
else
{a1[arr1[1]]+=(bd2-bd1);}
if((macha[1]+T12[arr1[1]])<=bd1 && machb[1]<=bd1||(macha[1]+T12[arr1[1]])>=bd2 &&
machb[1]>=bd2)
{b1[arr1[1]]=b1[arr1[1]];}
else
{b1[arr1[1]]+=(bd2-bd}
if((machb[1]+T23[arr1[1]])<=bd1 && machc[1]<=bd1||(machb[1]+T23[arr1[1]])>=bd2 &&
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machc[1]>=bd2)
{c1[arr1[1]]=c1[arr1[1]];}
else
{c1[arr1[1]]+=(bd2-bd1);}for(i=2;i<=n;i++)
{cout<<arr1[i]<<"t"<<macha11[i]<<"--"<<macha[i]<<" "<<"t"<<maxv1[i]<<"--"<<machb[i]<<"
"<<"t"<<maxv2[i]<<"--"<<machc[i]<<endl;
if(macha11[i]<=bd1 && macha[i]<=bd1 || macha11[i]>=bd2 && macha[i]>=bd2)
{a1[arr1[i]]=a1[arr1[i]];}
else
{a1[arr1[i]]+=(bd2-bd1);}
if(maxv1[i]<=bd1 && machb[i]<=bd1 || maxv1[i]>=bd2 && machb[i]>=bd2)
{b1[arr1[i]]=b1[arr1[i]];}
else
{b1[arr1[i]]+=(bd2-bd1);}
if(maxv2[i]<=bd1 && machc[i]<=bd1 || maxv2[i]>=bd2 && machc[i]>=bd2)
{c1[arr1[i]]=c1[arr1[i]];}
else
{c1[arr1[i]]+=(bd2-bd1);}}
cout<<"nnnTotal Elapsed Time (T) = "<<machc[n];
int j11[16];
for(i=1;i<=n;i++)
{ j11[i]=i;a11[arr1[i]]=a1[arr1[i]];b11[arr1[i]]=b1[arr1[i]];c11[arr1[i]]=c1[arr1[i]];}
cout<<endl<<"Modified Processing time after breakdown for the machines is:n";
cout<<"Jobs"<<"t"<<"Machine M1"<<"t"<<"t"<<"Machine M2" <<"t"<<"t"<<"Machine
M3"<<"t"<<"Weightage"<<endl;
for(i=1;i<=n;i++)
{cout<<endl;cout<<j11[i]<<"t"<<a11[i]<<"t"<<b11[i]<<"t"<<c11[i]<<"t"<<w[i];cout<<endl;}
float maxa12,minb12,minb22,maxc12;float g12[16],h12[16];
//Function for two ficticious machine G and H
//Findinglargest in a11
maxa12=a11[1]+T12[1]-s22[1];
for(i=2;i<n;i++)
{if((a11[i]+T12[i]-s22[i-1])>maxa12)maxa12=a11[i]+T12[i]-s22[i-1];}
//For finding smallest in b11
minb12=b11[1]+T23[1]-s33[1];
for(i=2;i<n;i++)
{if((b11[i]+T23[i]-s33[i])<minb12) minb12=b11[i]+T23[i]-s33[i];}
minb22=b11[1]+T12[1]-s11[i];
for(i=2;i<n;i++)
{if((b11[i]+T12[i]-s11[i])<minb22)minb22=b11[i]+T12[i]-s11[i];}
//Finding largest in c12
maxc12=c11[1]+T23[1]-s22[1];
for(i=2;i<n;i++)
{if((c11[i]+T23[i]-s22[i])>maxc12)maxc12=c11[i]+T23[i]-s22[i];}
if(maxa12>=minb22||maxc12>=minb12)
{g12[i]=fabs(a11[i]-T12[i]-b11[i]-T23[i]-maxs[i]);h12[i]=fabs(c11[i]-T12[i]-b11[i]-T23[i]+s33[i]);}
else
{cout<<"n data is not in Standard Form...nExitting";getch();exit(0);}
for(i=1;i<=n;i++)
{ g12[i]=fabs(a11[i]-T12[i]-b11[i]-T23[i]-maxs[i]); h12[i]=fabs(c11[i]-T12[i]-b11[i]-T23[i]+s33[i]);}
cout<<endl<<"Expected processing time for two fictious machines G and H: n";
for(i=1;i<=n;i++)
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{cout<<endl;cout<<j11[i]<<"t"<<g12[i]<<"t"<<h12[i]<<"t"<<w[i];cout<<endl;}
//To find minimum of G & H
float g11[16],h11[16];
for (i=1;i<=n;i++)
if(g12[i]<=h12[i])
{g11[i]=g12[i]+w[i];h11[i]=h12[i];}
else
{g11[i]=g12[i];h11[i]=h12[i]+w[i];}
float g21[16],h21[16];
for(i=1;i<=n;i++)
{g21[i]=g11[i]/w[i];h21[i]=h11[i]/w[i];}
for(i=1;i<=n;i++)
//calculate G_Beta and H_Beta
if(g21[group[1]]<h21[group[0]])
{minv1=g21[group[1]];}
else
{minv1=h21[group[0]];}
gbeta1=g21[group[0]]+g21[group[1]]-minv1;hbeta1=h21[group[0]]+h21[group[1]]-minv1;
cout<<endl<<endl<<"G_Beta1="<<gbeta1;cout<<endl<<"H_Beta1="<<hbeta1;
int j2[16];float g14[16],h14[16];int f1=1;
for(i=1;i<=n;i++)
{if(j11[i]==group[0]||j11[i]==group[1])
{f1--;}
else
{j2[f1]=j11[i];}f1++;}j2[n-1]=17;
for(i=1;i<=n-2;i++)
{g14[i]=g21[j2[i]];h14[i]=h21[j2[i]];}
g14[n-1]=gbeta1;h14[n-1]=hbeta1;
for(i=1;i<=n-1;i++)
float mingh1[16];char ch1[16];
for(i=1;i<=n-1;i++)
{if(g14[i]<h14[i]) {
mingh1[i]=g14[i];ch1[i]='g';}
else
{mingh1[i]=h14[i];ch1[i]='h'; }}
for(i=1;i<=n-1;i++)
{for(int j=1;j<=n-1;j++)
if(mingh1[i]<mingh1[j])
{float temp=mingh1[i]; int temp1=j2[i]; char d=ch1[i];mingh1[i]=mingh1[j]; j2[i]=j2[j];
ch1[i]=ch1[j];mingh1[j]=temp; j2[j]=temp1; ch1[j]=d;}}
// calculate beta scheduling
float sbeta1[16];int t2=1,s21=0;
for(i=1;i<=n-1;i++)
{if(ch1[i]=='h')
{ sbeta1[(n-s21-1)]=j2[i]; s21++;}
else if(ch1[i]=='g'){sbeta1[t2]=j2[i];t2++;}}
int arr2[16], m1=1;
cout<<endl<<endl<<"Job Scheduling:"<<"t";
29

Vol.1, No.3, 2011

for(i=1;i<=n-1;i++)
{if(sbeta1[i]==17)
{ arr2[m1]=group[0];arr2[m1+1]=group[1];cout<<group[0]<<" "<<group[1]<<"
";m1=m1+2;continue;}
else {cout<<sbeta1[i]<<" ";arr2[m1]=sbeta1[i];m1++;}}
//calculating total computation sequence;
float time1=0.0,macha12[16] ;float maxv11[16],maxv21[16];
macha1[1]=time1+a11[arr2[1]];
for(i=2;i<=n;i++)
{macha12[i]=macha1[i-1]+s11[arr2[i-1]];macha1[i]=macha12[i]+a11[arr2[i]];}
machb1[1]=macha1[1]+b11[arr2[1]]+T12[arr2[1]];
for(i=2;i<=n;i++)
{if((machb1[i-1]+s22[arr2[i-1]])>(macha1[i]+T12[arr2[i]])){maxv11[i]=machb1[i-1]+s22[arr2[i-1]];}
else {maxv11[i]=macha1[i]+T12[arr2[i]];}machb1[i]=maxv11[i]+b1[arr2[i]];}
machc1[1]=machb1[1]+c11[arr2[1]]+T23[arr2[1]];
for(i=2;i<=n;i++)
{if((machc1[i-1]+s33[arr2[i-1]])>(machb1[i]+T23[arr2[i]]))maxv21[i]=machc1[i-1]+s33[arr2[i-1]];
else maxv21[i]=machb1[i]+T23[arr2[i]];machc1[i]=maxv21[i]+c1[arr2[i]];}
float wft,sum1,sum2;sum2=0.0;
for(i=1;i<=n;i++)
{sum2=sum2+w[i];}
///displaying solution
cout<<"nnnnnttt #####THE SOLUTION##### ";
cout<<"nnt***************************************************************";
cout<<"nnnt Optimal Sequence is : ";
for(i=1;i<=n;i++)
{cout<<" "<<arr2[i];}
cout<<endl<<endl<<"In-Out Table is:"<<endl<<endl;
cout<<"Jobs"<<"t"<<"Machine M1"<<"t"<<"t"<<"Machine M2" <<"t"<<"t"<<"Machine M3"<<endl;
cout<<arr2[1]<<"t"<<time1<<"--"<<macha1[1]<<"
t"<<"t"<<macha1[1]+T12[arr2[1]]<<"--"<<machb1[1]<<"
t"<<"t"<<machb1[1]+T23[arr2[1]]<<"--"<<machc1[1]<<endl;sum1=0.0;
for(i=2;i<=n;i++)
{cout<<arr2[i]<<"t"<<macha12[i]<<"--"<<macha1[i]<<"
"<<"t"<<maxv11[i]<<"--"<<machb1[i]<<" "<<"t"<<maxv21[i]<<"--"<<machc1[i]<<endl;
sum1=sum1+(machc1[i]-macha12[i])*w[i]; }
cout<<"nnnTotal Elapsed Time (T) = "<<machc1[n];
wft=((machc1[1]*w[1])+sum1)/sum2;
cout<<"nnn The mean weighted flow time is = "<<wft;
cout<<"nnt***************************************************************";
getch();
}

7. Numerical Illustration
Consider 5 jobs, 3 machine flow shop problem with processing time, setup time each associated with their
respective probabilities, transportation time and weight of jobs are as given in table.
Jobs Machine M1 Machine M2 Machine M3 Weight
Ti ,1→2 Ti ,2→3 of jobs
i ai1 pi1 si1 qi1 ai2 pi2 si2 qi2 ai3 pi3 si3 qi3 wi

30

Vol.1, No.3, 2011

1 16 0.2 6 0.1 2 4 0.2 7 0.1 2 12 0.1 3 0.2 4
2 12 0.3 7 0.2 1 6 0.2 6 0.3 1 8 0.2 4 0.3 3
3 13 0.2 4 0.3 2 5 0.2 3 0.4 2 15 0.2 6 0.2 2
4 15 0.2 7 0.3 3 4 0.2 3 0.1 3 4 0.2 5 0.1 1
5 14 0.1 4 0.1 4 6 0.2 6 0.1 1 6 0.3 4 0.2 5
Table 2
Our objective is to obtain optimal or near optimal sequence when the break down interval is (a, b) = (10,15)
and jobs 2 & 5 are to be processed as equivalent job β = (2,5). Also calculate the total elapsed time and
mean weighted flow time.
Solution: As per Step 1: The expected processing times and expected setup times for machines M1, M2 and
M3 are as shown in table 3.
As per Step 2: Here, Max{Ai1 + Ti,1→2 – Si2} ≥ Min{Ai2 + Ti,1→2 – Si1}
Max{Ai3 + Ti,2→3 – Si2} ≥ Min{Ai2 + Ti,2→3 – Si3}; hence feasible solution.
As per Step. 3: The two fictitious machines G and H with processing times Gi and Hi are as shown in table
4.
As per Step 4 &5: The new reduced problem with processing time Gi'' and H i'' are as shown in table 5
As per step 6: The expected processing time of job block β(2,5) on fictitious machine G & H using
equivalent job block criteria given by Maggu & Das (1977) are
G " β = 0.466 + 1.08 − 1.08 = 0.466
H " β = 1.133 + 1.72 − 1.08 = 1.77
As per Step 8: The optimal sequence with minimum elapsed time using Johnson’s technique
is S = β – 4 – 3 – 1 i.e. 2 – 5 – 4 – 3 – 1
As per Step 9 & 10: The In-Out flow table for sequence S is as shown in table 6.
As per Step 11: On considering the effect of the break down interval (10 , 15) the original reduces to as
shown in table 7.
Now, On repeating the procedure to get the optimal sequence for the modified scheduling problem , we
have the sequence 3 – 2 – 5 – 4 – 1 which is optimal or near optimal. The In-Out flow table for the
modified scheduling problem is as shown in table 8.
The mean weighted flow time =
15.6 × 3 + (18.4 − 8.8 ) × 3 + ( 33.2 − 13.8 ) × 5 + ( 35.6 − 15.6 ) × 1 + ( 37.3 − 20.7 ) × 4
= 15.53
5 + 3 + 2 + 4 +1
Hence the total elapsed time is 37.3 hrs and the mean weighted flow time is 15.53 hrs.
Conclusion
The new method provides an optimal scheduling sequence with minimum total elapsed time whenever
mean weighted production flow time is taken into consideration for 3-machines, n-jobs flow shop
scheduling problems. This method is very easy to understand and will help the decision makers in
determining a best schedule for a given sets of jobs effectively to control job flow and provide a solution
for job sequencing. The study may further be extended by introducing the concept of Rental policy, due
date etc.
References
Baker, K. R.(1974), “Introduction of sequencing and scheduling”, John Wiley and Sons, New York.

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Bansal, S. P.(1986), “Resultant job in restricted two machine flow shop problem”, IJOMAS, 2(1986), 35-45.
Bellman, R.(1956), “Mathematical aspects of scheduling theory”, J. Soc. Indust. Appl. Math., 4(1956),
168-205.
Belwal & Mittal (2008), “n jobs machine flow shop scheduling problem with break down of machines,
transportation time and equivalent job block”, Bulletin of Pure & Applied Sciences-Mathematics, source
Vol. 27, Source Issue 1.
Chandramouli, A. B.(2005), “Heuristic approach for n-jobs, 3-machines flow-shop scheduling problem
involving transportation time, breakdown time and weights of jobs”, Mathematical and Computational
Applications,301-305.
Gupta, D. & Sharma, S. (2011), “Minimizing rental cost under specified rental policy in two stage flow
shop , the processing time associated with probabilities including breakdown interval and Job-block
criteria”, European Journal of Business and Management, 3(2),85-103.
Gupta, D., Sharma, S. & Seema (2011), “Heuristic approach for n-jobs, 3-machines flowshop scheduling
problem under specified rental policy, processing time associated with probabilities involving
transportation time, breakdown interval, Weightage of jobs and job block criteria ”, Mathematical Theory
and Modeling,1(1),30-36.
Johnson, S.M. (1954), “Optimal two and three stage production schedule with set up times included”,
Naval Research Logistics Quart.,1(1), 61-68.
Khodadadi, A.(2008), “Development of a new heuristic for three machines flow-shop scheduling problem
with transportation time of jobs”, World Applied Sciences Journal ,5(5), 598-601
Maggu, P.L. & Das G. (1977), “Equivalent jobs for job block in job scheduling”, Opsearch, 14(4),
277-281.
Miyazaki, S.& Nishiyama, N.(1980), “Analysis for minimizing weighted mean flow time in flow shop
scheduling”, J. O. R. Soc. Of Japan, 32 (1980), 118-132.
Nawaz, Ensore & Ham (1983), “A heuristic algorithm for the m-machine, n-job flow-shop sequencing
problem”, OMEGA, 91-95.
Narain, L. & Bagga, P.C.(1998), “Two machine flow shop problem with availability constraint on each
machine”, JISSOR, XXIV 1-4, 17-24.
Singh T.P.(1985), “On 2 x n flow-shop problems involving job-block, transportation time, arbitrary time
and break down machine time”, PAMS, XXI, No. 1—2 March.
Pandian, P. &. Rajendran, P. (2010), “Solving Constraint flow shop scheduling problems with three
machines”, Int. J. Contemp. Math. Sciences, 5(19), 921-929.
Yoshida & Hitomi (1979), “Optimal two stage production scheduling with set up times separated”, AIIE
Transactions, II, 261-263

Tables
Table 3: The expected processing times and expected setup times for machines M1, M2 and M3 are
Jobs Ai1 Si1 Ti ,1→2 Ai2 Si2 Ti ,2→3 Ai3 Si3 wi
1 3.2 0.6 2 0.8 0.7 2 1.2 0.6 4
2 3.6 1.4 1 1.2 1.8 1 1.6 1.2 3
3 2.6 1.2 2 1.0 1.2 2 3.0 1.2 2
4 3.0 2.1 3 0.8 0.3 3 0.8 0.5 1
5 1.4 0.4 4 1.2 0.6 1 1.8 0.8 5
Table 4: The two fictitious machines G and H with processing times Gi and Hi are
32

Vol.1, No.3, 2011

Jobs Gi Hi wi
1 2.3 3 4
2 1.4 0.4 3
3 3.6 0.8 2
4 5.9 5.5 1
5 5.4 3.6 5
Table 5: The new reduced problem with processing time Gi'' and H i'' are

Jobs Gi'' H i''
1 1.575 0.75
2 0.466 1.133
3 1.8 1.4
4 5.9 6.5
5 1.08 1.72
Table 6: The In-Out flow table for sequence S is

Jobs Machine M1 Ti ,1→2 Machine M2 Ti,2→3 Machine M3 wi
i In – Out In – Out In - Out
2 0 – 3.6 1 4.6 – 5.8 1 6.8 – 8.4 3
5 5 – 6.4 4 10.4 – 11.6 1 12.6 – 14.4 5
4 6.8 – 9.8 3 12.8 – 13.6 3 16.6 – 17.4 1
3 11.9 – 14.5 2 16.5 – 17.5 2 19.5 – 22.5 2
1 15.7 – 18.9 2 20.9 – 21.7 2 23.7 – 24.9 4
Table 7: On considering the effect of the break down interval (10, 15) the original reduces to

Jobs Ai1 Si1 Ti ,1→2 Ai2 Si2 Ti,2→3 Ai3 Si3 wi
1 3.2 0.6 2 0.8 0.7 2 1.2 0.6 4
2 3.6 1.4 1 1.2 1.8 1 1.6 1.2 3
3 7.6 1.2 2 1 1.2 2 3.0 1.2 2
4 3.0 2.1 3 5.8 0.3 3 0.8 0.5 1
5 1.4 0.4 4 6.2 0.6 1 6.8 0.8 5

33

Vol.1, No.3, 2011

Table 8: The In-Out flow table for the modified scheduling problem is

Jobs Machine M1 Ti ,1→2 Machine M2 Ti,2→3 Machine M3 wi
i In – Out In – Out In - Out
3 0.0 – 7.6 2 9.6 – 10.6 2 12.6 – 15.6 2
2 8.8 – 12.4 1 13.4 – 14.6 1 16.8 – 18.4 3
5 13.8 – 15.2 4 19.2 – 25.4 1 26.4 – 33.2 5
4 15.6 – 18.6 3 26.0 – 31.8 3 34.8 – 35.6 1
1 20.7 – 23.9 2 32.1 – 32.9 2 36.1 – 37.3 4

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